2

Let $E\subset [0,1]$ be a Lebesgue measurable set with measure $m(E)=1/2$.
Define the upper and lower densities \begin{align*} \theta_*(E,x) &= \liminf_{r\searrow 0} m(E\cap B_r(x))/2r\\ \theta^*(E,x) &= \limsup_{r\searrow 0} m(E\cap B_r(x))/2r, \end{align*} where $B_r(x) = (x-r,x+r)$. Does there necessarily exist a point $x\in I$ such that $$ 0 < \theta_*(E,x) \leq \theta^*(E,x) < 1? $$ Better yet, does there exist a constant $c>0$ independent of $E$ such that we can always find a point $x$ with $$ c \leq \theta_*(E,x) \leq \theta^*(E,x) \leq 1-c? $$

So far, I've tried defining $x=\inf\{y; \theta_*(E,y) = \theta^*(E,y) = 1\}$ in hopes of finding a "jump" on the left side, but in general it is possible for $\theta^*(E,x) = 0$. I've also been able to find a sequence $x_k, r_k$ such that $|E\cap B_{r_k}(x_k)|/2r_k = 1/2$ for all $k$, but this doesn't tell me anything about the density at $x = \lim x_k$.

felipeh
  • 3,790
  • An idea: Assume that all points have density $0$ or $1$. We can divide the interval in $N_1$ equal parts such that the approximate density on each is either $<2^{-4}$ or larger than $1-2^{-3}$. Imagine the intervals colored red or green according to the case. By Sperner's lemma there must be at least two consecutive interval of different colors. We look now in the interval formed by the union of these two. We repeat the same procedure with this smaller interval.

    This produces a sequence of nested intervals. The intersection of all of them gives a point. Look at that point.

     – OR. Mar 31 '15 at 17:26
  • Are you sure of the first point? My guess is that on most intervals we can ensure that the approximate density is either less than $2^{-4}$ or greather than $1-2^{-4}$, but perhaps not all of them. – felipeh Mar 31 '15 at 18:48
  • The set of points with density $0$ or $1$ has full measure. In particular it is dense. Each of them can be covered by an interval such that one of those inequalities holds. Those intervals cover $[0,1]$, and by compactness we can take finitely many of them that still cover $[0,1]$. (This argument is not perfect either. But more or less why I think it should be true). – OR. Mar 31 '15 at 18:51
  • I agree that you can cover $[0,1]$ with finitely many intervals on which one of the inequalities we want holds. My issue is that these intervals can have very different sizes. Consider for example a set $E$ that is a union of smaller and smaller dyadic intervals. – felipeh Mar 31 '15 at 19:06
  • See the following post: http://math.stackexchange.com/questions/693508/lebesgue-density-strictly-between-0-and-1?rq=1 –  Mar 31 '15 at 23:11
  • Thanks for the reference, that answers my question. – felipeh Apr 01 '15 at 01:15

1 Answers1

0

This is the answer posted here, pasted on this page for your convenience. I believe $A=E$ in your notation.

Since $\mu(A)>0$ and $\mu(\mathbb{R}\backslash A)>0$, there is a density point $y_1$ in $A$ and $z_1$ in $\mathbb{R}\backslash A$. Choose $r>0$ sufficiently small such that $$\frac{\mu(A\cap B_{r_1}(y_1))}{2r_1}> \frac{3}{4} \quad \text{and} \quad \frac{\mu(A\cap B_{r_1}(z_1))}{2r_1}< \frac{1}{4}.$$

Define $$f_r(x):= \frac{\mu(A \cap B_{r_1}(x))}{2r_1}.$$

For each $r>0$, $f_r$ is continuous, and so by Intermediate value theorem there is $x$ between $y_1$ and $z_1$ such that $$\frac{\mu(A \cap B_{r_1}(x))}{2r_1}=\frac{1}{2}.$$

Next, there are density points $y_2$ and $z_2$ of $A$ and $\mathbb{R}\backslash A$ in $B_{r_1}(x_1)$. Then choose $r_2<r_1/2$ sufficiently small such that $B_{r_2}(y_2), B_{r_2}(z_2) \subset B_{r_1}(x_1)$, and

$$\frac{\mu(A\cap B_{r_2}(y_2))}{2r_1}> \frac{3}{4} \quad \text{and} \quad \frac{\mu(A\cap B_{r_2}(z_2))}{2r_1}< \frac{1}{4}.$$

Again by Intermediate value theorem, there is $x_2$ between $y_2$ and $z_2$ such that $$\frac{\mu(A \cap B_{r_2}(x_2))}{2r_2}=\frac{1}{2}.$$

Continuing this, we have a sequence of points $\{x_n\}$ and radius $\{r_n\}$ such that $$\frac{\mu(A \cap B_{r_n}(x_n))}{2r_n}=\frac{1}{2}\quad \text{and} \quad B_{r_n}(x_n)\subset B_{r_{n-1}}(x_{n-1}) \quad \text{and} \quad r_n \downarrow 0,$$ and so $(x_n)$ converges to the point $x\in \bigcap_{n=1}^\infty \overline{B_{r_n}(x_n)}$.

Also, $x$ is not a density point of both $A$ and $\mathbb{R}\backslash A$ since $\mu(A\cap B_{2r_n}(x))\leq 2r_n+ 2r_n/2$ (since half of $B_{r_n}(x_n)$ is in $\mathbb{R}\backslash A$), and so $$ \frac{\mu(A\cap B_{2r_n}(x))}{4r_n}\leq \frac{3}{4} \text{ for each $n$},$$ which implies $x$ is not a density point of $A$. Similarly, $x$ is not a density point of $\mathbb{R}\backslash A$. This is the required $x$.

Community
  • 1
digiboy1
  • 1,614
  • 2
    This argument shows that there is a point $x$ which is not a density point of $A$ and not a density point of $\mathbb{R}\backslash A$. But we cannot conclude that $x$ has upper and lower densities away from $1$ and $0$ respectively, as was requested by the OP. – PatrickR Feb 25 '20 at 03:37